How do you write -3-2i32i in trigonometric form?

1 Answer
Apr 23, 2017

sqrt13(cos(2.55)-isin(2.55))13(cos(2.55)isin(2.55))

Explanation:

"to convert into "color(blue)"trigonometric form"to convert into trigonometric form

"that is " x+yitor(costheta+isintheta)that is x+yir(cosθ+isinθ) where

color(red)(bar(ul(|color(white)(2/2)color(black)(r=sqrt(x^2+y^2))color(white)(2/2)|)))

"and " color(red)(bar(ul(|color(white)(2/2)color(black)(theta=tan^-1(y/x);-pi< theta<=pi)color(white)(2/2)|)))

"here " x=-3" and " y=-2

rArrr=sqrt((-3)^2+(-2)^2)=sqrt13

"now " -3-2i" is in the 3rd quadrant so must ensure that"

theta" is in the 3rd quadrant"

theta=tan^-1(2/3)=0.588larrcolor(red)" related acute angle"

rArrtheta=-pi+0.588~~-2.55larrcolor(red)" in 3rd quadrant"

rArr-3-2i=sqrt13(cos(-2.55)+isin(-2.55))

• color(orange)" Reminder" cos(-theta)=costheta ; sin(-theta)=-sintheta

rArr-3-2i=sqrt13(cos(2.55)-isin(2.55))